Answer:
The length of the ladder is 7.21 feet
Step-by-step explanation:
Let
y = Height of the wall
x = Distance between the wall and the ladder (bottom)
L = Length of the ladder
So, the given parameters are:
[tex]\frac{dy}{dt} = -2ft/s[/tex]
When [tex]x = 4; \frac{dx}{dt} = 3ft/s[/tex]
The ladder, the wall and the ground forms a right-angled triangle where the hypotenuse is the length of the ladder; So:
[tex]L^2 = x^2 + y^2[/tex]
Differentiate with respect to time
[tex]2L\frac{dL}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}[/tex]
The length of the ladder does not change with time. So:
[tex]2L*0 = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}[/tex]
[tex]0 = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}[/tex]
Rewrite as:
[tex]2x\frac{dx}{dt} =- 2y\frac{dy}{dt}[/tex]
Divide both sides by 2
[tex]x\frac{dx}{dt} =- y\frac{dy}{dt}[/tex]
Recall that:
[tex]\frac{dy}{dt} = -2ft/s[/tex] and [tex]x = 4; \frac{dx}{dt} = 3ft/s[/tex]
So:
[tex]4 * 3 = -y * -2[/tex]
[tex]12 = 2y[/tex]
[tex]6 = y[/tex]
[tex]y = 6[/tex]
Substitute [tex]y = 6[/tex] and [tex]x =4[/tex] in: [tex]L^2 = x^2 + y^2[/tex]
[tex]L^2 =4^2 + 6^2[/tex]
[tex]L^2 =52[/tex]
[tex]L = \sqrt{52[/tex]
[tex]L = 7.21ft[/tex]
